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Class 12 Question Paper Alternative English Vibyor AHSEC 2017

Alternative English  Year - 2017 Full Marks : 100 

Class 12 Ncert Solution Maths|| 3d Geometry

Class 12 Ncert Solution Maths|| 3d Geometry

                                     Class 12 Ncert Solution Maths|| 3d Geometry

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Exercise 11.2

1. Show that the three lines with direction cosines
1213, 313, 413;  413, 1213, 313;  313, 413, 12131213, 313, 413;  413, 1213, 313;  313, 413, 1213 are mutually perpendicular.

Sol: Two lines are perpendicular if the sum of the product of their direction cosines is equal to zero.
If l1, m1, n1l1, m1, n1 and l2, m2, n2l2, m2, n2 be the direction cosines of two lines then they are perpendicular if       l1l2+m1m2+n1n2=0l1l2+m1m2+n1n2=0

l1l2+m1m2+n1n2
=1213.413+313.1213+413.313
= 481693616912169
=0
Thus the lines having direction cosines 1213, 313, 413;  413, 1213, 313 are perpendicular.

l1l2+m1m2+n1n2
=413.313+1213.413+313.1213
= 1216948169+36169
=0
Thus the lines having direction cosines 413, 1213, 313;  313, 413, 1213 are mutually perpendicular.

l1l2+m1m2+n1n2
=313.1213+413.313+413.1213
= 36169+1216948169
=0
The two lines having direction cosines 313, 413, 1213; 1213, 313, 413 are perpendicular.

Thus all the three lines are mutually perpendicular.

2. Show that the line through the points(1,-1,2), (3,4,-2) is perpendicular to the line through the point(0,3,2) and (3,5,6).

Sol: Direction ratio of the line passing through the points (1,-1,2), (3,4,-2) are 
(3-1),(4+1),(-2-2)
i.e. 2, 5, -4

Direction ratio of the line passing through the points (0, 3, 2), (3, 5, 6) are 
(3-0),(5-3),(6-2)
i.e. 3, 2, 4.

l1l2+m1m2+n1n2
=2.3 +5.2+(-4)4 
=6 + 10 -16
=0
Hence , the line through the points(1,-1,2), (3,4,-2) is perpendicular to the line through the point(0,3,2) and (1, 2, 5).

3. Show that the line through the points (4, 7, 8), (2, 3, 4) is parallel to the line through the points (-1, -2, 1), (3, 5, 6)

Sol:  Direction ratio of the line through (4, 7, 8), (2, 3, 4) is 
(2-4), (3-7), (4-8) 
i.e. (-2, -4, -4)

Direction ratio of the line through (-1, -2, 1), (1, 2, 5) is
(1+1), (2+2), (5-1)
i.e. 2, 4, 4

The direction ratios are proportional so it is clear that the lines are parallel.
Hence, the line through the points (4, 7, 8), (2, 3, 4) is parallel to the line through the points (-1, -2, 1), (3, 5, 6)

4. Find the equation of the line which passes through the point (1, 2, 3) and is parallel to the vector 3ˆi+2ˆj2ˆk

Sol: The position vector of the point  (1, 2, 3) is ˆi+2ˆj+3ˆk .Let it be denoted by a. Thus, aˆi+2ˆj+3ˆk

Let the other vector to which the line is parallel be denoted by b
Thus, b3ˆi+2ˆj2ˆk

The line which passes through the given point and parallel to b is given by 
r=a+λbr=(ˆi+2ˆj+3ˆk)+λ(3ˆi+2ˆj2ˆk)
this is the required equation of the line
6. Find the  cartesian equation of the line which passes through the point (-2,4,-5) and parallel to the line given by x+33=y45=z+86 

Sol: The line passing through a point x1,y1,z1 and having direction ratios a. b and c is given by
xx1a=yy1b=zz1c
The required line is parallel to the line  x+33=y45=z+86 (this line has direction ratios 3, 5 and 6)
Thus, the direction ratios of the required line is proportional to 3, 5 and 6.
Hence, the direction ratio of the required line is 3k, 5k and 6k.
Therefore , cartesian equation of the line which passes through the point (-2,4,-5) and parallel to the line given by x+33=y45=z+86 is

x(2)3k=y45k=z(5)6k
x+23k=y45k=z+56k
x+23=y45=z+56

7. The cartesian equation of aline is x53=y+47=z62

Sol: It is clear that the given line passes through the point (5, -4, 6).
Position vector of the point (5, -4, 6) is a=5ˆi4ˆj+6ˆk
From the equation of the line itcan be seen that its direction ratios are 3, 7 and 2.
Thus the line parallel to the given line is b=3ˆi+7ˆj+2ˆk
Hence the equation of the line in vector form is
r=5ˆi4ˆj+6ˆk+λ(3ˆi+7ˆj+2ˆk)

8. Find the vector and the cartesian equations of the line that passes through the origin and  (5, -2, 3).

Sol: The required line passes through (0,0,0) and (5,-2,3). So the direction ratios of this line are(5-0),(-2-0),(3-0) i.e 5, -2, 3.
Hence the equation of the line passing through (0,0,0) and having direction ratio 5, -2, 3 is given by
x05=y02=z03

x5=y2=z3

The position vector for the point (0,0,0) is a=0 and the position vector for the point (5,-2, 3) is b=5ˆi2ˆj+3ˆk
The equation of a line in vector form  is given by
r=a+λ(b
 r=0+λ(5ˆi2ˆj+3ˆk
r=λ(5ˆi2ˆj+3ˆk)


9. Find the vector and cartesian equations of the line that passes through the points (3, -2, -5). (3, -2, 6)

Sol:  The required line passes through  (3, -2, -5) and (3, -2, 6). So the direction ratios of this line are(3-3),(-2-(-2)),(6-(-5)) i.e 0, 0, 11.
Hence the equation of the line passing through (3, -2, -5) and having direction ratio 0, 0, 11 is given by
x30=y(2)0=z(5)11

x30=y+20=z+511

The position vector for the point (3, -2, -5) is a=3ˆi2ˆj5ˆk and the position vector for the point (0, 0 , 11) is b=11ˆk
The equation of a line in vector form  is given by
r=a+λb
 r=3ˆi2ˆj5ˆk+λ(11ˆk)

10. Find the angle between the following pair of lines:
(i) r=2ˆi5ˆj+ˆk+λ(3ˆi+2ˆj+6ˆk) and
r=7ˆi6ˆk+μ(ˆi+2ˆj+2ˆk)

(ii) r=3ˆi+ˆj2ˆk+λ(ˆiˆj2ˆk) and
r=2ˆiˆj56ˆk+μ(3ˆi5ˆj4ˆk)

Sol: Angle between two lines is given by
cosθ=|b1.b2|b1|.|b2||
(i)
Given that b1=3ˆi+2ˆj+6ˆk and b2=ˆi+2ˆj+2ˆk
|b1| = 32+22+62=49=7
and |b2| = 12+22+22=9=3

Hence, cosθ=|(3ˆi+2ˆj+6ˆk).(ˆi+2ˆj+2ˆk)7.3|
= |3+4+1221|=1921
Thus,   θ=cos1(1921)

(ii) Given that b1=ˆiˆj2ˆk and b2=3ˆi5ˆj4ˆk
|b1| = 12+12+22=6
and |b2| = 32+52+42=50=52

Hence, cosθ=|(ˆiˆj2ˆk).(3ˆi5ˆj4ˆk)6.52|

= |3+5+8103|=16103=853

Thus,   θ=cos1(83)


11. Find the angle between the following pair of lines:
(i)  x22=y15=z+33 and x+21=y48=z54

(ii) x2=y2=z1 and x54=y21=z38

Sol:  The angle between two lines having direction ratios a1,b1,c1 and a2, b2 and c2 is given by cosθ=|a1a2+b1b2+c1c2a21+b21+c21a22+b22+c22|

(i)  The angle between the given two lines is given by 

cosθ=|2(1)+5.8+4(3)22+52+(3)2(1)2+82+42|

= |2+40124+25+91+64+16|

=263881

θ=cos1(26938)

(ii)  The angle between the given two lines is given by 

cosθ=|2.4+2.1+1.822+22+1242+12+82|

= |8+2+84+4+116+2+64|

=18981

=183×9

θ=cos1(23)

12. Find the values of p so that the lines 1x3=7y142p=z32 and 
77x3p=y51=6z5 are at right angles.

Sol:  The given two lines will be at right angles when the sum of the product of the direction ratios is equal to zero. i.e a1a2+b1b2+c1c2=0
 (3)3p7+2p71+2×(5)=0

9p7+2p710=0

11p=10

p=1011

13.  Show that the lines x57=y+25=z1 and x1=y2=z3 are perpendicular to each other.

Sol:   The given two lines will be perpendiculkar to each other when the sum of the product of the direction ratios is equal to zero. i.e a1a2+b1b2+c1c2=0
L.H.S = 7×1+(5)×2+1×3
=7-10+3
=0 =R.H.S
Since the sum is equal to zero so the lines are perpendicular.



Class 12 Ncert Solution Maths|| 3d Geometry
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